What is the integral of e^{3x}(e^{3x-8})^5 ?

The integral of e^{3x}(e^{3x-8})^5 is 1/18 e^{18x-40}+C

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integral of e^{3x}(e^{3x-8})^5

\int e^{3 x} (e^{3 x - 8})^{5} d x

\frac{1}{18} e^{18 x - 40} + C

Solution steps

\int e^{3 x} (e^{3 x - 8})^{5} d x

Simplify

e^{3 x} (e^{3 x - 8})^{5} : e^{18 x - 40}

= \int e^{18 x - 40} d x

Apply u-substitution

= \int e^{u} \frac{1}{18} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{18} e^{u}

Substitute back

u = 18 x - 40

= \frac{1}{18} e^{18 x - 40}

Add a constant to the solution

= \frac{1}{18} e^{18 x - 40} + C

What is the integral of e^{3x}(e^{3x-8})^5 ?
The integral of e^{3x}(e^{3x-8})^5 is 1/18 e^{18x-40}+C